2.1 The Electromagnetic Model Equations
25
Inverting matrix equation (2.28) gives
δxδyδz
⎡
⎢
⎢
⎢
⎣
ρ klm
ρ k+1lm
ρ kl+1m
ρ k+1l+1m
⎤
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
x 11 y 01 y 10
D
−y 11 x 01 x 10
D
y 01 y 11 x 10
D
−y 01 x 11 x 10
D
−y 00 y 11 x 01
D
y 11 x 01 x 00
D
−y 01 y 11 x 00
D
x 00 x 11 y 01
D
−x 11 y 00 y 10
D
x 00 x 11 y 10
D
−y 00 y 11 x 10
D
y 00 x 11 x 10
D
x 01 y 00 y 10
D
−x 00 x 01 y 10
D
x 00 y 01 y 10
D
−y 00 x 01 x 10
D
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
E x
klm
E
y
klm
E x
kl+1m
E
y
k+1lm
⎤
⎥
⎥
⎥
⎦
,
(2.35)
where D = x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 . The solution for the σ ’s is
σ klm =
(x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 ) δxδyδz
x 11 y 01 y 10 E x
klm − y 11 x 01 x 10 E
y
klm + y 01 y 11 x 10 E x
kl+1m − y 01 x 11 x 10 E
y
k+1lm
σ k+1lm =
(x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 ) δxδyδz
−y 00 x 01 y 11 E x
klm + y 11 x 01 x 00 E
y
klm − y 01 y 11 x 00 E x
kl+1m + y 01 x 11 x 00 E
y
k+1lm
σ kl+1m =
(x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 ) δxδyδz
−x 11 y 00 y 10 E x
klm + x 11 x 00 y 10 E
y
klm − y 00 y 11 x 10 E x
kl+1m + y 00 x 11 x 10 E
y
k+1lm
σ k+1l+1m =
(x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 ) δxδyδz
x 01 y 00 y 10 E x
klm − y 10 x 01 x 00 E
y
klm + y 01 y 10 x 00 E x
kl+1m − y 00 x 01 x 10 E
y
k+1lm
.
(2.36)
This solution for the σ ’s will not exist unless the system, (2.28), is independent.
There are many situations that are likely to arise which will result in dependence
among these equations, and we, therefore, turn to a different method of ‘solving’
(2.28), which we call the ‘two-cell’ hypothesis. Consider a 2 × 2 array of adjacent
cells in the x- and y-directions. We call this array a ‘window pane,’ and ask for the
solution of (2.28) under the hypothesis that two adjacent cells in the window pane
have equal conductivities. The answer is:
σ klm = σ k+1lm =
(x 00 + x 10 )
E x
klm
σ klm = σ kl+1m =
(y 00 + y 01 )
E
y
klm
σ kl+1m = σ k+1l+1m =
(x 01 + x 11 )
E x
kl+1m
σ k+1lm = σ k+1l+1m =
(y 10 + y 11 )
E
y
k+1lm
(2.37)
25
Inverting matrix equation (2.28) gives
δxδyδz
⎡
⎢
⎢
⎢
⎣
ρ klm
ρ k+1lm
ρ kl+1m
ρ k+1l+1m
⎤
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
x 11 y 01 y 10
D
−y 11 x 01 x 10
D
y 01 y 11 x 10
D
−y 01 x 11 x 10
D
−y 00 y 11 x 01
D
y 11 x 01 x 00
D
−y 01 y 11 x 00
D
x 00 x 11 y 01
D
−x 11 y 00 y 10
D
x 00 x 11 y 10
D
−y 00 y 11 x 10
D
y 00 x 11 x 10
D
x 01 y 00 y 10
D
−x 00 x 01 y 10
D
x 00 y 01 y 10
D
−y 00 x 01 x 10
D
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
E x
klm
E
y
klm
E x
kl+1m
E
y
k+1lm
⎤
⎥
⎥
⎥
⎦
,
(2.35)
where D = x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 . The solution for the σ ’s is
σ klm =
(x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 ) δxδyδz
x 11 y 01 y 10 E x
klm − y 11 x 01 x 10 E
y
klm + y 01 y 11 x 10 E x
kl+1m − y 01 x 11 x 10 E
y
k+1lm
σ k+1lm =
(x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 ) δxδyδz
−y 00 x 01 y 11 E x
klm + y 11 x 01 x 00 E
y
klm − y 01 y 11 x 00 E x
kl+1m + y 01 x 11 x 00 E
y
k+1lm
σ kl+1m =
(x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 ) δxδyδz
−x 11 y 00 y 10 E x
klm + x 11 x 00 y 10 E
y
klm − y 00 y 11 x 10 E x
kl+1m + y 00 x 11 x 10 E
y
k+1lm
σ k+1l+1m =
(x 00 x 11 y 01 y 10 − y 00 y 11 x 01 x 10 ) δxδyδz
x 01 y 00 y 10 E x
klm − y 10 x 01 x 00 E
y
klm + y 01 y 10 x 00 E x
kl+1m − y 00 x 01 x 10 E
y
k+1lm
.
(2.36)
This solution for the σ ’s will not exist unless the system, (2.28), is independent.
There are many situations that are likely to arise which will result in dependence
among these equations, and we, therefore, turn to a different method of ‘solving’
(2.28), which we call the ‘two-cell’ hypothesis. Consider a 2 × 2 array of adjacent
cells in the x- and y-directions. We call this array a ‘window pane,’ and ask for the
solution of (2.28) under the hypothesis that two adjacent cells in the window pane
have equal conductivities. The answer is:
σ klm = σ k+1lm =
(x 00 + x 10 )
E x
klm
σ klm = σ kl+1m =
(y 00 + y 01 )
E
y
klm
σ kl+1m = σ k+1l+1m =
(x 01 + x 11 )
E x
kl+1m
σ k+1lm = σ k+1l+1m =
(y 10 + y 11 )
E
y
k+1lm
(2.37)